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Question

Physics Question on Gravitational Potential Energy

A particle of mass mm is moved from the surface of the earth to a height hh. The work done by an external agency to do this is (1) mghm g h for h<<Rh < < R (3) 12mgh\frac{1}{2} m g h for h=Rh=R (2) mghm g h for all hh (4) 12mgh-\frac{1}{2} m g h for h=Rh=R

A

1, 2 and 3 are correct

B

1 and 2 are correct

C

2 and 4 are correct

D

1 and 3 are correct

Answer

1 and 3 are correct

Explanation

Solution

Ui=GMmRU_{i}=-\frac{G M m}{R} Uf=GMm(R+h)U_{f}=-\frac{G M m}{(R+h)} Work done == change in gravitational potential energy E=UfUiE=U_{f}-U_{i} =GMm[1R+h1R]=-G M m\left[\frac{1}{R+h}-\frac{1}{R}\right] W=GMmR[(1+hR)11]W=-\frac{G M m}{R}\left[\left(1+\frac{h}{R}\right)^{-1}-1\right] W=GMmR[1hR1]W=-\frac{G M m}{R}\left[1-\frac{h}{R}-1\right] W=GMmhR2W=\frac{G M m h}{R^{2}} W=mghW=m g h for hRh \leq R Also Ui=GMmRU_{i}=-\frac{G M m}{R} and on the surface h=Rh=R Uf=GMmR+RU_{f}=-\frac{G M m}{R+R} =GMm2R=-\frac{G M m}{2 R} Work done (W)=UfUi(W)=U_{f}-U_{i} =GMm2R=\frac{G M m}{2 R} W=12mgRW=\frac{1}{2} m g R [g=GMR2]\left[\because g=\frac{G M}{R^{2}}\right]